9.6: Chi-Square Tests
In this section, we will study a number of important hypothesis tests that fall under the general term chi-square tests . These are named, as you might guess, because in each case the test statistics has (in the limit) a chi-square distribution. Although there are several different tests in this general category, they all share some common themes:
- In each test, there are one or more underlying multinomial samples, Of course, the multinomial model includes the Bernoulli model as a special case.
- Each test works by comparing the observed frequencies of the various outcomes with expected frequencies under the null hypothesis.
- If the model is incompletely specified , some of the expected frequencies must be estimated; this reduces the degrees of freedom in the limiting chi-square distribution.
We will start with the simplest case, where the derivation is the most straightforward; in fact this test is equivalent to a test we have already studied. We then move to successively more complicated models.
The One-Sample Bernoulli Model
Suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample from the Bernoulli distribution with unknown success parameter \(p \in (0, 1)\). Thus, these are independent random variables taking the values 1 and 0 with probabilities \(p\) and \(1 - p\) respectively. We want to test \(H_0: p = p_0\) versus \(H_1: p \ne p_0\), where \(p_0 \in (0, 1)\) is specified. Of course, we have already studied such tests in the Bernoulli model. But keep in mind that our methods in this section will generalize to a variety of new models that we have not yet studied.
Let \(O_1 = \sum_{j=1}^n X_j\) and \(O_0 = n - O_1 = \sum_{j=1}^n (1 - X_j)\). These statistics give the number of times (frequency) that outcomes 1 and 0 occur, respectively. Moreover, we know that each has a binomial distribution; \(O_1\) has parameters \(n\) and \(p\), while \(O_0\) has parameters \(n\) and \(1 - p\). In particular, \(\E(O_1) = n p\), \(\E(O_0) = n (1 - p)\), and \(\var(O_1) = \var(O_0) = n p (1 - p)\). Moreover, recall that \(O_1\) is sufficient for \(p\). Thus, any good test statistic should be a function of \(O_1\). Next, recall that when \(n\) is large, the distribution of \(O_1\) is approximately normal, by the central limit theorem. Let \[ Z = \frac{O_1 - n p_0}{\sqrt{n p_0 (1 - p_0)}} \] Note that \(Z\) is the standard score of \(O_1\) under \(H_0\). Hence if \(n\) is large, \(Z\) has approximately the standard normal distribution under \(H_0\), and therefore \(V = Z^2\) has approximately the chi-square distribution with 1 degree of freedom under \(H_0\). As usual, let \(\chi_k^2\) denote the quantile function of the chi-square distribution with \(k\) degrees of freedom.
An approximate test of \(H_0\) versus \(H_1\) at the \(\alpha\) level of significance is to reject \(H_0\) if and only if \(V \gt \chi_1^2(1 - \alpha)\).
The test above is equivalent to the unbiased test with test statistic \(Z\) (the approximate normal test) derived in the section on Tests in the Bernoulli model.
For purposes of generalization, the critical result in the next exercise is a special representation of \(V\). Let \(e_0 = n (1 - p_0)\) and \(e_1 = n p_0\). Note that these are the expected frequencies of the outcomes 0 and 1, respectively, under \(H_0\).
\(V\) can be written in terms of the observed and expected frequencies as follows: \[ V = \frac{(O_0 - e_0)^2}{e_0} + \frac{(O_1 - e_1)^2}{e_1} \]
This representation shows that our test statistic \(V\) measures the discrepancy between the expected frequencies, under \(H_0\), and the observed frequencies. Of course, large values of \(V\) are evidence in favor of \(H_1\). Finally, note that although there are two terms in the expansion of \(V\) in Exercise 3, there is only one degree of freedom since \(O_0 + O_1 = n\). The observed and expected frequencies could be stored in a \(1 \times 2\) table.
The Multi-Sample Bernoulli Model
Suppose now that we have samples from several (possibly) different, independent Bernoulli trials processes. Specifically, suppose that \(\bs{X}_i = (X_{i,1}, X_{i,2}, \ldots, X_{i,n_i})\) is a random sample of size \(n_i\) from the Bernoulli distribution with unknown success parameter \(p_i \in (0, 1)\) for each \(i \in \{1, 2, \ldots, m\}\). Moreover, the samples \((\bs{X}_1, \bs{X}_2, \ldots, \bs{X}_m)\) are independent. We want to test hypotheses about the unknown parameter vector \(\bs{p} = (p_1, p_2, \ldots, p_m)\). There are two common cases that we consider below, but first let's set up the essential notation that we will need for both cases. For \(i \in \{1, 2, \ldots, m\}\) and \(j \in \{0, 1\}\), let \(O_{i,j}\) denote the number of times that outcome \(j\) occurs in sample \(\bs{X}_i\). The observed frequency \(O_{i,j}\) has a binomial distribution; \(O_{i,1}\) has parameters \(n_i\) and \(p_i\) while \(O_{i,0}\) has parameters \(n_i\) and \(1 - p_i\).
The Completely Specified Case
Consider a specified parameter vector \(\bs{p}_0 = (p_{0,1}, p_{0,2}, \ldots, p_{0,m}) \in (0, 1)^m\). We want to test the null hypothesis \(H_0: \bs{p} = \bs{p}_0\), versus \(H_1: \bs{p} \ne \bs{p}_0\). Since the null hypothesis specifies the value of \(p_i\) for each \(i\), this is called the completely specified case . Now let \(e_{i,0} = n_i (1 - p_{i,0})\) and let \(e_{i,1} = n_i p_{i,0}\). These are the expected frequencies of the outcomes 0 and 1, respectively, from sample \(\bs{X}_i\) under \(H_0\).
If \(n_i\) is large for each \(i\), then under \(H_0\) the following test statistic has approximately the chi-square distribution with \(m\) degrees of freedom: \[ V = \sum_{i=1}^m \sum_{j=0}^1 \frac{(O_{i,j} - e_{i,j})^2}{e_{i,j}} \]
Proof
This follows from the result above and independence.
As a rule of thumb,
large
means that we need \(e_{i,j} \ge 5\) for each \(i \in \{1, 2, \ldots, m\}\) and \(j \in \{0, 1\}\). But of course, the larger these expected frequencies the better.
Under the large sample assumption, an approximate test of \(H_0\) versus \(H_1\) at the \(\alpha\) level of significance is to reject \(H_0\) if and only if \(V \gt \chi_m^2(1 - \alpha)\).
Once again, note that the test statistic \(V\) measures the discrepancy between the expected and observed frequencies, over all outcomes and all samples. There are \(2 \, m\) terms in the expansion of \(V\) in Exercise 4, but only \(m\) degrees of freedom, since \(O_{i,0} + O_{i,1} = n_i\) for each \(i \in \{1, 2, \ldots, m\}\). The observed and expected frequencies could be stored in an \(m \times 2\) table.
The Equal Probability Case
Suppose now that we want to test the null hypothesis \(H_0: p_1 = p_2 = \cdots = p_m\) that all of the success probabilities are the same, versus the complementary alternative hypothesis \(H_1\) that the probabilities are not all the same. Note, in contrast to the previous model, that the null hypothesis does not specify the value of the common success probability \(p\). But note also that under the null hypothesis, the \(m\) samples can be combined to form one large sample of Bernoulli trials with success probability \(p\). Thus, a natural approach is to estimate \(p\) and then define the test statistic that measures the discrepancy between the expected and observed frequencies, just as before. The challenge will be to find the distribution of the test statistic.
Let \(n = \sum_{i=1}^m n_i\) denote the total sample size when the samples are combined. Then the overall sample mean, which in this context is the overall sample proportion of successes, is \[ P = \frac{1}{n} \sum_{i=1}^m \sum_{j=1}^{n_i} X_{i,j} = \frac{1}{n} \sum_{i=1}^m O_{i,1} \] The sample proportion \(P\) is the best estimate of \(p\), in just about any sense of the word. Next, let \(E_{i,0} = n_i \, (1 - P)\) and \(E_{i,1} = n_i \, P\). These are the estimated expected frequencies of 0 and 1, respectively, from sample \(\bs{X}_i\) under \(H_0\). Of course these estimated frequencies are now statistics (and hence random) rather than parameters. Just as before, we define our test statistic \[ V = \sum_{i=1}^m \sum_{j=0}^1 \frac{(O_{i,j} - E_{i,j})^2}{E_{i,j}} \] It turns out that under \(H_0\), the distribution of \(V\) converges to the chi-square distribution with \(m - 1\) degrees of freedom as \(n \to \infty\).
An approximate test of \(H_0\) versus \(H_1\) at the \(\alpha\) level of significance is to reject \(H_0\) if and only if \(V \gt \chi_{m-1}^2(1 - \alpha)\).
Intuitively, we lost a degree of freedom over the completely specified case because we had to estimate the unknown common success probability \(p\). Again, the observed and expected frequencies could be stored in an \(m \times 2\) table.
The One-Sample Multinomial Model
Our next model generalizes the one-sample Bernoulli model in a different direction. Suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a sequence of multinomial trials. Thus, these are independent, identically distributed random variables, each taking values in a set \(S\) with \(k\) elements. If we want, we can assume that \(S = \{0, 1, \ldots, k - 1\}\); the one-sample Bernoulli model then corresponds to \(k = 2\). Let \(f\) denote the common probability density function of the sample variables on \(S\), so that \(f(j) = \P(X_i = j)\) for \(i \in \{1, 2, \ldots, n\}\) and \(j \in S\). The values of \(f\) are assumed unknown, but of course we must have \(\sum_{j \in S} f(j) = 1\), so there are really only \(k - 1\) unknown parameters. For a given probability density function \(f_0\) on \(S\) we want to test \(H_0: f = f_0\) versus \(H_1: f \ne f_0\).
By this time, our general approach should be clear. We let \(O_j\) denote the number of times that outcome \(j \in S\) occurs in sample \(\bs{X}\): \[ O_j = \sum_{i=1}^n \bs{1}(X_i = j) \] Note that \(O_j\) has the binomial distribution with parameters \(n\) and \(f(j)\). Thus, \(e_j = n \, f_0(j)\) is the expected number of times that outcome \(j\) occurs, under \(H_0\). Out test statistic, of course, is \[ V = \sum_{j \in S} \frac{(O_j - e_j)^2}{e^j} \] It turns out that under \(H_0\), the distribution of \(V\) converges to the chi-square distribution with \(k - 1\) degrees of freedom as \(n \to \infty\). Note that there are \(k\) terms in the expansion of \(V\), but only \(k - 1\) degrees of freedom since \(\sum_{j \in S} O_j = n\).
An approximate test of \(H_0\) versus \(H_1\) at the \(\alpha\) level of significance is to reject \(H_0\) if and only if \(V \gt \chi_{k-1}^2(1 - \alpha)\).
Again, as a rule of thumb, we need \(e_j \ge 5\) for each \(j \in S\), but the larger the expected frequencies the better.
The Multi-Sample Multinomial Model
As you might guess, our final generalization is to the multi-sample multinomial model. Specifically, suppose that \(\bs{X}_i = (X_{i,1}, X_{i,2}, \ldots, X_{i,n_i})\) is a random sample of size \(n_i\) from a distribution on a set \(S\) with \(k\) elements, for each \(i \in \{1, 2, \ldots, m\}\). Moreover, we assume that the samples \((\bs{X}_1, \bs{X}_2, \ldots, \bs{X}_m)\) are independent. Again there is no loss in generality if we take \(S = \{0, 1, \ldots, k - 1\}\). Then \(k = 2\) reduces to the multi-sample Bernoulli model , and \(m = 1\) corresponds to the one-sample multinomial model .
Let \(f_i\) denote the common probability density function of the variables in sample \(\bs{X}_i\), so that \(f_i(j) = \P(X_{i,l} = j)\) for \(i \in \{1, 2, \ldots, m\}\), \(l \in \{1, 2, \ldots, n_i\}\), and \(j \in S\). These are generally unknown, so that our vector of parameters is the vector of probability density functions: \(\bs{f} = (f_1, f_2, \ldots, f_m)\). Of course, \(\sum_{j \in S} f_i(j) = 1\) for \(i \in \{1, 2, \ldots, m\}\), so there are actually \(m \, (k - 1)\) unknown parameters. We are interested in testing hypotheses about \(\bs{f}\). As in the multi-sample Bernoulli model, there are two common cases that we consider below, but first let's set up the essential notation that we will need for both cases. For \(i \in \{1, 2, \ldots, m\}\) and \(j \in S\), let \(O_{i,j}\) denote the number of times that outcome \(j\) occurs in sample \(\bs{X}_i\). The observed frequency \(O_{i,j}\) has a binomial distribution with parameters \(n_i\) and \(f_i(j)\).
The Completely Specified Case
Consider a given vector of probability density functions on \(S\), denoted \(\bs{f}_0 = (f_{0,1}, f_{0,2}, \ldots, f_{0,m})\). We want to test the null hypothesis \(H_0: \bs{f} = \bs{f}_0\), versus \(H_1: \bs{f} \ne \bs{f}_0\). Since the null hypothesis specifies the value of \(f_i(j)\) for each \(i\) and \(j\), this is called the completely specified case . Let \(e_{i,j} = n_i \, f_{0,i}(j)\). This is the expected frequency of outcome \(j\) in sample \(\bs{X}_i\) under \(H_0\).
If \(n_i\) is large for each \(i\), then under \(H_0\), the test statistic \(V\) below has approximately the chi-square distribution with \(m \, (k - 1)\) degrees of freedom: \[ V = \sum_{i=1}^m \sum_{j \in S} \frac{(O_{i,j} - e_{i,j})^2}{e_{i,j}} \]
Proof
This follows from the one-sample multinomial case and independence.
As usual, our rule of thumb is that we need \(e_{i,j} \ge 5\) for each \(i \in \{1, 2, \ldots, m\}\) and \(j \in S\). But of course, the larger these expected frequencies the better.
Under the large sample assumption, an approximate test of \(H_0\) versus \(H_1\) at the \(\alpha\) level of significance is to reject \(H_0\) if and only if \(V \gt \chi_{m \, (k - 1)}^2(1 - \alpha)\).
As always, the test statistic \(V\) measures the discrepancy between the expected and observed frequencies, over all outcomes and all samples. There are \(m k\) terms in the expansion of \(V\) in Exercise 8, but we lose \(m\) degrees of freedom, since \(\sum_{j \in S} O_{i,j} = n_i\) for each \(i \in \{1, 2, \ldots, m\}\).
The Equal PDF Case
Suppose now that we want to test the null hypothesis \(H_0: f_1 = f_2 = \cdots = f_m\) that all of the probability density functions are the same, versus the complementary alternative hypothesis \(H_1\) that the probability density functions are not all the same. Note, in contrast to the previous model, that the null hypothesis does not specify the value of the common success probability density function \(f\). But note also that under the null hypothesis, the \(m\) samples can be combined to form one large sample of multinomial trials with probability density function \(f\). Thus, a natural approach is to estimate the values of \(f\) and then define the test statistic that measures the discrepancy between the expected and observed frequencies, just as before.
Let \(n = \sum_{i=1}^m n_i\) denote the total sample size when the samples are combined. Under \(H_0\), our best estimate of \(f(j)\) is \[ P_j = \frac{1}{n} \sum_{i=1}^m O_{i,j} \] Hence our estimate of the expected frequency of outcome \(j\) in sample \(\bs{X}_i\) under \(H_0\) is \(E_{i,j} = n_i P_j\). Again, this estimated frequency is now a statistic (and hence random) rather than a parameter. Just as before, we define our test statistic \[ V = \sum_{i=1}^m \sum_{j \in S} \frac{(O_{i,j} - E_{i,j})^2}{E_{i,j}} \] As you no doubt expect by now, it turns out that under \(H_0\), the distribution of \(V\) converges to a chi-square distribution as \(n \to \infty\). But let's see if we can determine the degrees of freedom heuristically.
The limiting distribution of \(V\) has \((k - 1) (m - 1)\) degrees of freedom.
Proof
There are \(k \, m\) terms in the expansion of \(V\). We lose \(m\) degrees of freedom since \(\sum_{j \in S} O_{i,j} = n_i\) for each \(i \in \{1, 2, \ldots, m\}\). We must estimate all but one of the probabilities \(f(j)\) for \(j \in S\), thus losing \(k - 1\) degrees of freedom.
An approximate test of \(H_0\) versus \(H_1\) at the \(\alpha\) level of significance is to reject \(H_0\) if and only if \(V \gt \chi_{(k - 1) \, (m - 1)}^2(1 -\alpha)\).
A Goodness of Fit Test
A goodness of fit test is an hypothesis test that an unknown sampling distribution is a particular, specified distribution or belongs to a parametric family of distributions. Such tests are clearly fundamental and important. The one-sample multinomial model leads to a quite general goodness of fit test.
To set the stage, suppose that we have an observable random variable \(X\) for an experiment, taking values in a general set \(S\). Random variable \(X\) might have a continuous or discrete distribution, and might be single-variable or multi-variable. We want to test the null hypothesis that \(X\) has a given, completely specified distribution, or that the distribution of \(X\) belongs to a particular parametric family.
Our first step, in either case, is to sample from the distribution of \(X\) to obtain a sequence of independent, identically distributed variables \(\bs{X} = (X_1, X_2, \ldots, X_n)\). Next, we select \(k \in \N_+\) and partition \(S\) into \(k\) (disjoint) subsets. We will denote the partition by \(\{A_j: j \in J\}\) where \(\#(J) = k\). Next, we define the sequence of random variables \(\bs{Y} = (Y_1, Y_2, \ldots, Y_n)\) by \(Y_i = j\) if and only if \(X_i \in A_j\) for \(i \in \{1, 2, \ldots, n\}\) and \(j \in J\).
\(\bs{Y}\) is a multinomial trials sequence with parameters \(n\) and \(f\), where \(f(j) = \P(X \in A_j)\) for \(j \in J\).
The Completely Specified Case
Let \(H\) denote the statement that \(X\) has a given, completely specified distribution. Let \(f_0\) denote the probability density function on \(J\) defined by \(f_0(j) = \P(X \in A_j \mid H)\) for \(j \in J\). To test hypothesis \(H\), we can formally test \(H_0: f = f_0\) versus \(H_1: f \ne f_0\), which of course, is precisely the problem we solved in the one-sample multinomial model .
Generally, we would partition the space \(S\) into as many subsets as possible, subject to the restriction that the expected frequencies all be at least 5.
The Partially Specified Case
Often we don't really want to test whether \(X\) has a completely specified distribution (such as the normal distribution with mean 5 and variance 9), but rather whether the distribution of \(X\) belongs to a specified parametric family (such as the normal). A natural course of action in this case would be to estimate the unknown parameters and then proceed just as above. As we have seen before, the expected frequencies would be statistics \(E_j\) because they would be based on the estimated parameters. As a rule of thumb, we lose a degree of freedom in the chi-square statistic \(V\) for each parameter that we estimate, although the precise mathematics can be complicated.
A Test of Independence
Suppose that we have observable random variables \(X\) and \(Y\) for an experiment, where \(X\) takes values in a set \(S\) with \(k\) elements, and \(Y\) takes values in a set \(T\) with \(m\) elements. Let \(f\) denote the joint probability density function of \((X, Y)\), so that \(f(i, j) = \P(X = i, Y = j)\) for \(i \in S\) and \(j \in T\). Recall that the marginal probability density functions of \(X\) and \(Y\) are the functions \(g\) and \(h\) respectively, where \begin{align} g(i) = & \sum_{j \in T} f(i, j), \quad i \in S \\ h(j) = & \sum_{i \in S} f(i, j), \quad j \in T \end{align} Usually, of course, \(f\), \(g\), and \(h\) are unknown. In this section, we are interested in testing whether \(X\) and \(Y\) are independent, a basic and important test. Formally then we want to test the null hypothesis \[ H_0: f(i, j) = g(i) \, h(j), \quad (i, j) \in S \times T \] versus the complementary alternative \(H_1\).
Our first step, of course, is to draw a random sample \((\bs{X}, \bs{Y}) = ((X_1, Y_1), (X_2, Y_2), \ldots, (X_n, Y_n))\) from the distribution of \((X, Y)\). Since the state spaces are finite, this sample forms a sequence of multinomial trials. Thus, with our usual notation, let \(O_{i,j}\) denote the number of times that \((i, j)\) occurs in the sample, for each \((i, j) \in S \times T\). This statistic has the binomial distribution with trial parameter \(n\) and success parameter \(f(i, j)\). Under \(H_0\), the success parameter is \(g(i) \, h(j)\). However, since we don't know the success parameters, we must estimate them in order to compute the expected frequencies. Our best estimate of \(f(i, j)\) is the sample proportion \(\frac{1}{n} O_{i,j}\). Thus, our best estimates of \(g(i)\) and \(h(j)\) are \(\frac{1}{n} N_i\) and \(\frac{1}{n} M_j\), respectively, where \(N_i\) is the number of times that \(i\) occurs in sample \(\bs{X}\) and \(M_j\) is the number of times that \(j\) occurs in sample \(\bs{Y}\): \begin{align} N_i & = \sum_{j \in T} O_{i,j} \\ M_j & = \sum_{i \in S} O_{i,j} \end{align} Thus, our estimate of the expected frequency of \((i, j)\) under \(H_0\) is \[ E_{i,j} = n \, \frac{1}{n} \, N_i \frac{1}{n} \, M_j = \frac{1}{n} \, N_i \, M_j \] Of course, we define our test statistic by \[ V = \sum_{i \in J} \sum_{j \in T} \frac{(O_{i,j} - E_{i,j})^2}{E_{i,j}} \] As you now expect, the distribution of \(V\) converges to a chi-square distribution as \(n \to \infty\). But let's see if we can determine the appropriate degrees of freedom on heuristic grounds.
The limiting distribution of \(V\) has \((k - 1) \, (m - 1)\) degrees of freedom.
Proof
There are \(k m\) terms in the expansion of \(V\). We lose one degree of freedom since \(\sum_{i \in S} \sum_{j \in T} O_{i,j} = n\). We must estimate all but one of the probabilities \(g(i)\) for \(i \in S\), thus losing \(k - 1\) degrees of freedom. We must estimate all but one of the probabilities \(h(j)\) for \(j \in T\), thus losing \(m - 1\) degrees of freedom.
An approximate test of \(H_0\) versus \(H_1\) at the \(\alpha\) level of significance is to reject \(H_0\) if and only if \(V \gt \chi_{(k-1) (m-1)}^2(1 - \alpha)\).
The observed frequencies are often recorded in a \(k \times m\) table, known as a contingency table , so that \(O_{i,j}\) is the number in row \(i\) and column \(j\). In this setting, note that \(N_i\) is the sum of the frequencies in the \(i\)th row and \(M_j\) is the sum of the frequencies in the \(j\)th column. Also, for historical reasons, the random variables \(X\) and \(Y\) are sometimes called factors and the possible values of the variables categories .
Computational and Simulation Exercises
Computational Exercises
In each of the following exercises, specify the number of degrees of freedom of the chi-square statistic, give the value of the statistic and compute the \(P\)-value of the test.
A coin is tossed 100 times, resulting in 55 heads. Test the null hypothesis that the coin is fair.
Answer
1 degree of freedom, \(V = 1\), \(P = 0.3173\).
Suppose that we have 3 coins. The coins are tossed, yielding the data in the following table:
| Heads | Tails | |
|---|---|---|
| Coin 1 | 29 | 21 |
| Coin 2 | 23 | 17 |
| Coin 3 | 42 | 18 |
- Test the null hypothesis that all 3 coin are fair.
- Test the null hypothesis that coin 1 has probability of heads \(\frac{3}{5}\); coin 2 is fair; and coin 3 has probability of heads \(\frac{2}{3}\).
- Test the null hypothesis that the 3 coins have the same probability of heads.
Answer
- 3 degree of freedom, \(V = 11.78\), \(P = 0.008\).
- 3 degree of freedom, \(V = 1.283\), \(P = 0.733\).
- 2 degree of freedom, \(V = 2.301\), \(P = 0.316\).
A die is thrown 240 times, yielding the data in the following table:
| Score | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Frequency | 57 | 39 | 28 | 28 | 36 | 52 |
- Test the null hypothesis that the die is fair.
- Test the null hypothesis that the die is an ace-six flat die (faces 1 and 6 have probability \(\frac{1}{4}\) each while faces 2, 3, 4, and 5 have probability \(\frac{1}{8}\) each).
Answer
- 5 degree of freedom, \(V = 18.45\), \(P = 0.0024\).
- 5 degree of freedom, \(V = 5.383\), \(P = 0.3709\).
Two dice are thrown, yielding the data in the following table:
| Score | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Die 1 | 22 | 17 | 22 | 13 | 22 | 24 |
| Die 2 | 44 | 24 | 19 | 19 | 18 | 36 |
- Test the null hypothesis that die 1 is fair and die 2 is an ace-six flat.
- Test the null hypothesis that all the dice have have the same probability distribuiton.
Answer
- 10 degree of freedom, \(V = 6.2\), \(P = 0.798\).
- 5 degree of freedom, \(V = 7.103\), \(P = 0.213\).
A university classifies faculty by rank as instructors , assistant professors , associate professors , and full professors . The data, by faculty rank and gender, are given in the following contingency table. Test to see if faculty rank and gender are independent.
| Faculty | Instructor | Assistant Professor | Associate Professor | Full Professor |
|---|---|---|---|---|
| Male | 62 | 238 | 185 | 115 |
| Female | 118 | 122 | 123 | 37 |
Answer
3 degrees of freedom, \(V = 70.111\), \(P \approx 0\).
Data Analysis Exercises
The Buffon trial data set gives the results of 104 repetitions of Buffon's needle experiment. The number of crack crossings is 56. In theory, this data set should correspond to 104 Bernoulli trials with success probability \(p = \frac{2}{\pi}\). Test to see if this is reasonable.
Answer
1 degree of freedom, \(V = 4.332\), \(P = 0.037\).
Test to see if the alpha emissions data come from a Poisson distribution.
Answer
We partition of \(\N\) into 17 subsets: \(\{0, 1\}\), \(\{x\}\) for \(x \in \{2, 3, \ldots, 16\}\), and \(\{17, 18, \ldots \}\). There are 15 degrees of freedom. The estimated Poisson parameter is 8.367, \(V = 9.644\), \(P = 0.842\).
Test to see if Michelson's velocity of light data come from a normal distribution.
Answer
Using the following partition of \(\R\): \(\{(-\infty, 750), [750, 775), [775, 800), [800, 825), [825, 850), [850, 875), [875, 900), [900, 925), [925, 950), [950, 975), [975, \infty)\}\). We have 8 degrees of freedom, \(V = 11.443\), \(P = 0.178\).
Simulation Exercises
In the simulation exercises below, you will be able to explore the goodness of fit test empirically.
In the dice goodness of fit experiment, set the sampling distribution to fair, the sample size to 50, and the significance level to 0.1. Set the test distribution as indicated below and in each case, run the simulation 1000 times. In case (a), give the empirical estimate of the significance level of the test and compare with 0.1. In the other cases, give the empirical estimate of the power of the test. Rank the distributions in (b)-(d) in increasing order of apparent power. Do your results seem reasonable?
- fair
- ace-six flats
- the symmetric, unimodal distribution
- the distribution skewed right
In the dice goodness of fit experiment, set the sampling distribution to ace-six flats, the sample size to 50, and the significance level to 0.1. Set the test distribution as indicated below and in each case, run the simulation 1000 times. In case (a), give the empirical estimate of the significance level of the test and compare with 0.1. In the other cases, give the empirical estimate of the power of the test. Rank the distributions in (b)-(d) in increasing order of apparent power. Do your results seem reasonable?
- fair
- ace-six flats
- the symmetric, unimodal distribution
- the distribution skewed right
In the dice goodness of fit experiment, set the sampling distribution to the symmetric, unimodal distribution, the sample size to 50, and the significance level to 0.1. Set the test distribution as indicated below and in each case, run the simulation 1000 times. In case (a), give the empirical estimate of the significance level of the test and compare with 0.1. In the other cases, give the empirical estimate of the power of the test. Rank the distributions in (b)-(d) in increasing order of apparent power. Do your results seem reasonable?
- the symmetric, unimodal distribution
- fair
- ace-six flats
- the distribution skewed right
In the dice goodness of fit experiment, set the sampling distribution to the distribution skewed right, the sample size to 50, and the significance level to 0.1. Set the test distribution as indicated below and in each case, run the simulation 1000 times. In case (a), give the empirical estimate of the significance level of the test and compare with 0.1. In the other cases, give the empirical estimate of the power of the test. Rank the distributions in (b)-(d) in increasing order of apparent power. Do your results seem reasonable?
- the distribution skewed right
- fair
- ace-six flats
- the symmetric, unimodal distribution
Suppose that \(D_1\) and \(D_2\) are different distributions. Is the power of the test with sampling distribution \(D_1\) and test distribution \(D_2\) the same as the power of the test with sampling distribution \(D_2\) and test distribution \(D_1\)? Make a conjecture based on your results in the previous three exercises.
In the dice goodness of fit experiment, set the sampling and test distributions to fair and the significance level to 0.05. Run the experiment 1000 times for each of the following sample sizes. In each case, give the empirical estimate of the significance level and compare with 0.05.
- \(n = 10\)
- \(n = 20\)
- \(n = 40\)
- \(n = 100\)
In the dice goodness of fit experiment, set the sampling distribution to fair, the test distributions to ace-six flats, and the significance level to 0.05. Run the experiment 1000 times for each of the following sample sizes. In each case, give the empirical estimate of the power of the test. Do the powers seem to be converging?
- \(n = 10\)
- \(n = 20\)
- \(n = 40\)
- \(n = 100\)


17.5: Activities
In each of the following, describe the requested combination of relations in words (i.e. in the form “a is related to b if …”). Try to “simplify” your description, if possible.
In Task h and Task i, the symbol \(\equiv_k\) represents a relation on \(\mathbb{Z}\text{,}\) where \(m \equiv_k n\) means that \(m\) and \(n\) have the same remainder when divided by \(k\text{.}\) (It may help to know that this is equivalent to \(k\) dividing the difference \(m - n\text{.}\))
- \(< \cup >\) on \(\mathbb{Z}\text{.}\)
- Union of “longer than” and “shorter than” on \(\Sigma^\ast\) for some alphabet \(\Sigma\text{.}\)
- Union of “longer than”, “shorter than”, and “same length as” on \(\Sigma^\ast\) for some alphabet \(\Sigma\text{.}\)
- Intersection of “longer than” and “shorter than” on \(\Sigma^\ast\) for some alphabet \(\Sigma\text{.}\)
- The complement of \(\le\) on \(\mathbb{Z}\text{.}\)
- The inverse of \(\le\) on \(\mathbb{Z}\text{.}\)
- The inverse of “\(x R y\) if \(2 x + 3 y = 0\)” on \(\mathbb{Z}\text{.}\)
- The intersection of \(\equiv_5\) and \(\equiv_7\) on \(\mathbb{Z}\text{.}\)
- The intersection of \(\equiv_2\) and \(\equiv_4\) on \(\mathbb{Z}\text{.}\)
In each of the following, you are given a set \(A\) and a relation \(R\) on \(A\text{.}\) Determine which of the properties reflexive , symmetric , antisymmetric , and transitive \(R\) possesses.
- \(A = \mathbb{Z}\text{,}\) \(R\) is \(<\text{.}\)
- \(A\) is the set of all straight lines in the plane, \(R\) means “is parallel to.”
- \(A\) is the set of all straight lines in the plane, \(R\) means “is perpendicular to.”
- \(A = \Sigma^\ast\) for some alphabet \(\Sigma\text{,}\) \(R\) means “is the same length as.”
- \(A = \Sigma^\ast\) for some alphabet \(\Sigma\text{,}\) \(R\) means “is shorter than.”
- \(A = \Sigma^\ast\) for some alphabet \(\Sigma\text{,}\) \(x\) is some fixed choice of letter in \(\Sigma\text{,}\) \(R\) means “contains the same number of occurrences of \(x\) as.”
- \(A\) is an arbitrary set, \(R\) is the empty relation .
- \(A\) is an arbitrary set, \(R\) is the universal relation .
- Suppose \(R\) is a relation on a set \(A\text{.}\) Convince yourself that \(R \cup R^{-1}\) is symmetric. (See the Symmetric Relation Test .)
- Recall that \(\vert\) represents the relation “divides” on sets of integers. Draw the directed graph for \(\vert\) on the set \(A = \{2,4,6,8,10,12,14,16\}\text{.}\) Then describe how to obtain the graph for the symmetric relation \(\vert \cup \vert ^{-1}\) as an undirected graph from the graph of \(R\) using only an eraser.
For each of the properties reflexive , symmetric , antisymmetric , and transitive , carry out the following.
Assume that \(R\) and \(S\) are nonempty relations on a set \(A\) that both have the property. For each of \(R^C\text{,}\) \(R \cup S\text{,}\) \(R \cap S\text{,}\) and \(R^{-1}\text{,}\) determine whether the new relation
- must also have that property;
- might have that property, but might not; or
- cannot have that property.
Any time you answer Statement i or Statement iii, outline a proof. Any time you answer Statement ii, provide two examples: one where the new relation has the property, and one where the new relation does not. (You may use graphs to describe your examples.)


10.7: The Effect of Temperature Changes on Equilibrium
When temperature is the stress that affects a system at equilibrium, there are two important consequences:
- an increase in temperature will favor that reaction direction that absorbs heat (i.e. the endothermic reaction)
- the value of K eq will change
Consider the following equilibrium system
\[\ce{N_2O_4(g) \leftrightarrow 2NO_2(g)}\]
with \(\ce{\Delta H^{\circ}={58.0}\:kJ}\)
Delta H is a measurement showing heat was gained by the reaction. We will cover heat and energy soon.
We see by the sign of ΔH° that the forward reaction is endothermic. Heat is absorbed (required as a reactant) when the reaction proceeds as
\[\ce{N_2O_4(g) \rightarrow 2NO_2 (g)}\]
By adding more heat, equilibrium will shift to use up the additional heat, thus favoring this forward direction.
Why will K eq change, when it did not change when concentration, pressure, and volume were the applied stresses?
When temperature changes cause an equilibrium to shift, one entire side of the reaction equation is favored over the other side. Mathematically, this will alter the value of K eq as follows:
\[\ce{K_{eq}=\dfrac{[products]}{[reactants]}}\]
|
if the forward reaction is favored |
more products are produced; fewer reactants |
K eq will increase |
|
if the reverse reaction is favored |
fewer products; more reactants |
K eq will decrease |
So in our example given above, increasing the temperature will favor the forward direction. The value of K eq will increase. Removing heat (making the system colder) will favor the exothermic reaction - the exothermic reaction releases heat to the surroundings, thus "replacing" the heat that has been removed.


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5.9: Thinking about the Material
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Reflect and research
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What was the name of the publication in which Newton’s published his three laws, and when was it published?
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When did Galileo come up with his principle of inertia?
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Suppose that Newton grew up in an accelerating train, with no knowledge that he is living in an accelerating train. What would Newton’s first law look like in this world?
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When you skate on ice, there is kinetic friction between your skates and the ice. Does the coefficient of kinetic friction depend on the temperature of the ice? If yes, what is the optimal temperature for skating with the least amount of friction?
To try at home
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Place two books stacked on each other on the palm of one hand held horizontally. Use your other hand to press down (and forward) on the top book and try to move the bottom book. No matter how hard you push down (to increase the force of friction between the two books), you cannot make the bottom one move. How come?
To try in the lab
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Propose an experiment to determine whether gravitational and inertial mass are equal.
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Propose an experiment to measure the coefficients of static and kinetic friction between a block and a surface.


10.3: Defining Equivalent Ratios
Lesson
Let's investigate equivalent ratios some more.
Exercise \(\PageIndex{1}\): Dots and Half Dots
Dot Pattern 1:
Dot Pattern 2:
Exercise \(\PageIndex{2}\): Tuna Casserole
Here is a recipe for tuna casserole.
Ingredients
- 3 cups cooked elbow-shaped pasta
- 6 ounce can tuna, drained
- 10 ounce can cream of chicken soup
- 1 cup shredded cheddar cheese
- cups French fried onions
Instructions
Combine the pasta, tuna, soup, and half of the cheese. Transfer into a 9 inch by 18 inch baking dish. Put the remaining cheese on top. Bake 30 minutes at 350 degrees. During the last 5 minutes, add the French fried onions. Let sit for 10 minutes before serving.
- What is the ratio of the ounces of soup to the cups of shredded cheese to the cups of pasta in one batch of casserole?
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How much of each of these 3 ingredients would be needed to make:
- twice the amount of casserole?
- half the amount of casserole?
- five times the amount of casserole?
- one-fifth the amount of casserole?
- What is the ratio of cups of pasta to ounces of tuna in one batch of casserole?
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How many batches of casserole would you make if you used the following amounts of ingredients?
- 9 cups of pasta and 18 ounces of tuna?
- 36 ounces of tuna and 18 cups of pasta?
- 1 cup of pasta and 2 ounces of tuna?
Are you ready for more?
The recipe says to use a 9 inch by 18 inch baking dish. Determine the length and width of a baking dish with the same height that could hold:
- Twice the amount of casserole
- Half the amount of casserole
- Five times the amount of casserole
- One-fifth the amount of casserole
Exercise \(\PageIndex{3}\): What Are Equivalent Ratios?
The ratios \(5:3\) and \(10:6\) are equivalent ratios .
- Is the ratio \(15:12\) equivalent to these? Explain your reasoning.
- Is the ratio \(30:18\) equivalent to these? Explain your reasoning.
- Give two more examples of ratios that are equivalent to \(5:3\).
- How do you know when ratios are equivalent and when they are not equivalent?
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Write a definition of
equivalent ratios
.
Pause here so your teacher can review your work and assign you a ratio to use for your visual display. -
Create a visual display that includes:
- the title “Equivalent Ratios”
- your best definition of equivalent ratios
- the ratio your teacher assigned to you
- at least two examples of ratios that are equivalent to your assigned ratio
- an explanation of how you know these examples are equivalent
- at least one example of a ratio that is not equivalent to your assigned ratio
- an explanation of how you know this example is not equivalent
Be prepared to share your display with the class.
Summary
All ratios that are equivalent to \(a:b\) can be made by multiplying both \(a\) and \(b\) by the same number.
For example, the ratio \(18:12\) is equivalent to \(9:6\) because both 9 and 6 are multiplied by the same number: 2.
\(3:2\) is also equivalent to \(9:6\), because both 9 and 6 are multiplied by the same number: \(\frac{1}{3}\).
Is \(18:15\) equivalent to \(9:6\)?
No, because 18 is \(9\cdot 2\), but 15 is not \(6\cdot 2\).
Glossary Entries
Definition: Equivalent Ratios
Two ratios are equivalent if you can multiply each of the numbers in the first ratio by the same factor to get the numbers in the second ratio. For example, \(8:6\) is equivalent to \(4:3\), because \(8\cdot\frac{1}{2}=4\) and \(6\cdot\frac{1}{2}=3\).
A recipe for lemonade says to use 8 cups of water and 6 lemons. If we use 4 cups of water and 3 lemons, it will make half as much lemonade. Both recipes taste the same, because and are equivalent ratios.
| cups of water | number of lemons |
|---|---|
| \(8\) | \(6\) |
| \(4\) | \(3\) |
Practice
Exercise \(\PageIndex{4}\)
Each of these is a pair of equivalent ratios. For each pair, explain why they are equivalent ratios or draw a diagram that shows why they are equivalent ratios.
- \(4:5\) and \(8:10\)
- \(18:3\) and \(6:1\)
- \(2:7\) and \(10,000:35,000\)
Exercise \(\PageIndex{5}\)
Explain why \(6:4\) and \(18:8\) are not equivalent ratios.
Exercise \(\PageIndex{6}\)
Are the ratios \(3:6\) and \(6:3\) equivalent? Why or why not?
Exercise \(\PageIndex{7}\)
This diagram represents 3 batches of light yellow paint. Draw a diagram that represents 1 batch of the same shade of light yellow paint.
(From Unit 2.2.2)
Exercise \(\PageIndex{8}\)
In the fruit bowl there are 6 bananas, 4 apples, and 3 oranges.
- For every 4 __________________, there are 3 __________________.
- The ratio of __________________ to __________________ is \(6:3\).
- The ratio of __________________ to __________________ is 4 to 6.
- For every 1 orange, there are ______ bananas.
(From Unit 2.1.1)
Exercise \(\PageIndex{9}\)
Write fractions for points \(A\) and \(B\) on the number line.
(From Unit 2.1.1)


9.8: Discussion- Practice Paraphrasing
Academic writing also involves heavy use of paraphrasing sources. Paraphrasing is actually much more common than quoting, particularly in APA-style writing.
Paraphrasing has several advantages:
- it lets you keep a consistent tone and voice throughout the essay
- it demonstrates your mastery of the concepts coming from outside sources
- it lets you be flexible in wording and vocabulary to best meet the needs of your readers
Paraphrasing seems simple on the surface: it’s just putting another author’s ideas into your own original words. In practice, though, this is one of the most challenging aspects of writing academic work.
To help us all feel more comfortable and confident with our paraphrasing skills, let’s practice it here.
In your post, copy and paste the original wording (a direct quote) from the source you’re using for the Source Evaluation Essay. Be sure to include the title of the source and a link to it, if possible, and put the quote inside of quotation marks.
Beneath this quote, draft a paraphrase that states the idea of the quotation in unique language. The paraphrase should include a “signal” phrase, so that we have some context for where it’s coming from. Guidance about how to draft a paraphrase can be found in earlier module contents.
Your post should be about 100-200 words. It doesn’t have to be grammatically perfect, but should use standard English (no text-speak, please) and normal capitalization rules.
You will also need to return to this Discussion to reply to at least two of your classmates’ posts. Content could include, but is not limited to, any of the following: Commenting on the style or quality of your classmate’s paraphrase. Be sure to point out if the paraphrase contains too much borrowed language from the original article that your classmate may not have noticed.
Responses are weighed as heavily as your initial posting, and should be roughly as long (100-200 words) when combined. Responses should indicate you’ve read your classmate’s post carefully. Include specific details from the post you’re responding to in your reply.
Contributors and Attributions
- Composition II. Authored by : Alexis McMillan-Clifton. Provided by : Tacoma Community College. Located at : http://www.tacomacc.edu . Project : Kaleidoscope Open Course Initiative. License : CC BY: Attribution


10.1: Introduction
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The 2010 Patient Protection and Affordable Care Act (ACA), also known as Obamacare, represented a substantial overhaul of the U.S. healthcare system. 1 Given its potential impact, interest group representatives (lobbyists) from the insurance industry, hospitals, medical device manufacturers, and organizations representing doctors, patients, and employers all tried to influence what the law would look like and the way it would operate. Ordinary people took to the streets to voice their opinion (Figure 10.1). Some state governors sued to prevent a requirement in the law that their states expand Medicaid coverage. A number of interest groups challenged the law in court, where two Supreme Court decisions have left it largely intact.
Interest groups like those for and against the ACA play a fundamental role in representing individuals, corporate interests, and the public before the government. They help inform the public and lawmakers about issues, monitor government actions, and promote policies that benefit their interests, using all three branches of government at the federal, state, and local levels.
In this chapter, we answer several key questions about interest groups. What are they, and why and how do they form? How do they provide avenues for political participation? Why are some groups advantaged by the lobbying of government representatives, while others are disadvantaged? Finally, how do interest groups try to achieve their objectives, and how are they regulated?


13.5: Summary
13.1 Prenatal Testing during the First Trimester
Many prenatal tests are recommended during the first trimester of pregnancy. Nurses are responsible for obtaining a comprehensive obstetric and menstrual history and educating pregnant persons on the routine, screening, and diagnostic tests available and the reasons they are performed and recommended. The nurse may also assist with some of the procedures and can help prepare the pregnant person by explaining what to expect during the procedure. It is important for nurses to educate the pregnant person on the screenings recommended and inform them that the screenings predict the risk of the fetus having a genetic, structural, or neural tube defect. Additional diagnostic testing is needed if the screenings come back abnormal.
Nurses also use therapeutic communication with pregnant persons and their support persons. It is important for the nurse to support the pregnant person in any decisions they make concerning their pregnancy. Part of this support includes referrals to specialists in maternal-fetal medicine, genetics, and infectious disease.
13.2 Prenatal Testing during the Second Trimester
Prenatal testing in the second trimester includes screening and diagnostic tests. Some of the testing is performed after an abnormal first trimester screening result. Some of the prenatal testing allows for interventions to take place during the procedure. Before the second trimester screening or diagnostic testing, the nurse will collect any important information that may be helpful to the health-care provider when interpreting results, such as gravida, para, living children, gestation in weeks, and any maternal health problems.
The nurse reinforces education about the risks and benefits of prenatal testing, the test results, and recommendations made by the health-care provider. The nurse has a very important role in educating the pregnant person on the screenings and procedures that are recommended, answering any questions, preparing the pregnant person for what to expect during the procedures, and educating the pregnant person on what the results mean and any additional recommendations. The nurse should provide the pregnant person and their partner with clear explanations of what the tests assess, the purpose of the tests, and the frequency of any tests recommended. It will be important for the pregnant person to know how long the test will take and what happens during the test to help reduce any anxiety. The pregnant person may need additional support depending on the results of the test performed, and additional testing or referrals may be appropriate. Abnormal results can often cause anxiety in the pregnant person.
13.3 Prenatal Testing during the Third Trimester
Routine prenatal screening is usually complete prior to the third trimester. However, additional screenings and tests may be recommended. If results from routine screening performed in the second trimester are inconclusive or positive, additional testing may be needed in the third trimester. Additionally, if any complications occur in the third trimester, additional fetal surveillance may be needed to check on the well-being of the fetus.
The nurse has an important role in educating and supporting the pregnant person on all the recommended screening and diagnostic tests performed in the third trimester. These nursing actions assist the pregnant person in their decisions on what prenatal testing is right for them and their fetus.


8.4: Colligative Properties- Boiling Point Elevation and Freezing Point Depression
Make sure you thoroughly understand the following essential ideas:
- Explain why boiling point elevation follows naturally from Raoult's law.
- Explaining freezing point depression is admittedly a bit more difficult, but you should nevertheless be able to explain how the application of salt on an ice-covered road can cause the ice to melt.
The colligative properties really depend on the escaping tendency of solvent molecules from the liquid phase. You will recall that the vapor pressure is a direct measure of escaping tendency, so we can use these terms more or less interchangeably.
Boiling Point Elevation
If addition of a nonvolatile solute lowers the vapor pressure of the solution via Raoult's law , then it follows that the temperature must be raised to restore the vapor pressure to the value corresponding to the pure solvent. In particular, the temperature at which the vapor pressure is 1 atm will be higher than the normal boiling point by an amount known as the boiling point elevation . The exact relation between the boiling point of the solution and the mole fraction of the solvent is rather complicated, but for dilute solutions the elevation of the boiling point is directly proportional to the molal concentration of the solute:
Bear in mind that the proportionality constant K B is a property of the solvent because this is the only component that contributes to the vapor pressure in the model we are considering in this section.
| solvent | normal bp, °C | K b , K mol –1 kg |
|---|---|---|
| water | 100 | 0.514 |
| ethanol | 79 | 1.19 |
| acetic acid | 118 | 2.93 |
| carbon tetrachloride | 76.5 | 5.03 |
Sucrose (C 22 O 11 H 22 , 342 g mol –1 ), like many sugars, is highly soluble in water; almost 2000 g will dissolve in 1 L of water, giving rise to what amounts to pancake syrup. Estimate the boiling point of such a sugar solution.
Solution
moles of sucrose:
\[ \dfrac{2000\, g}{342\, g\, mol^{–1}} = 5.8\; mol\]
mass of water: assume 1000 g (we must know the density of the solution to find its exact value)
The molality of the solution is (5.8 mol) ÷ (1.0 kg) = 5.8 m.
Using the value of K b from the table, the boiling point will be raised by (0.514 K mol –1 kg) × (5.8 mol kg –1 ) = 3.0 K, so the boiling point will be 103° C.
Freezing Point Depression
The freezing point of a substance is the temperature at which the solid and liquid forms can coexist indefinitely — that is, they are in equilibrium. Under these conditions molecules pass between the two phases at equal rates because their escaping tendencies from the two phases are identical. Suppose that a liquid solvent and its solid (water and ice, for example) are in equilibrium ( below), and we add a non-volatile solute (such as salt, sugar, or automotive antifreeze liquid) to the water. This will have the effect of reducing the mole fraction of H 2 O molecules in the liquid phase, and thus reduce the tendency of these molecules to escape from it, not only into the vapor phase (as we saw above), but also into the solid (ice) phase. This will have no effect on the rate at which H 2 O molecules escape from the ice into the water phase, so the system will no longer be in equilibrium and the ice will begin to melt .
If we wish to keep the solid from melting, the escaping tendency of molecules from the solid must be reduced. This can be accomplished by reducing the temperature ; this lowers the escaping tendency of molecules from both phases, but it affects those in the solid more than those in the liquid, so we eventually reach the new, lower freezing point where the two quantities are again in exact balance and both phases can coexist .
If you prefer to think in terms of vapor pressures, you can use the same argument if you bear in mind that the vapor pressures of the solid and liquid must be the same at the freezing point. Dilution of the liquid (the solvent) by the nonvolatile solute reduces the vapor pressure of the solvent according to Raoult’s law, thus reducing the temperature at which the vapor pressures of the liquid and frozen forms of the solution will be equal. As with boiling point elevation, in dilute solutions there is a simple linear relation between the freezing point depression and the molality of the solute:
\[ \Delta T_f = K_f \dfrac{\text{moles of solute}}{\text{kg of solvent}}\]
Note that K f values are all negative !
| Solvent | Normal Freezing Point (°C) | K f (K mol –1 kg) |
|---|---|---|
| water | 0.0 | –1.86 |
| acetic acid | 16.7 | –3.90 |
| benzene | 5.5 | –5.10 |
| camphor | 180 | –40.0 |
| cyclohexane | 6.5 | –20.2 |
| phenol | 40 | –7.3 |
The use of salt to de-ice roads is a common application of this principle. The solution formed when some of the salt dissolves in the moist ice reduces the freezing point of the ice. If the freezing point falls below the ambient temperature, the ice melts. In very cold weather, the ambient temperature may be below that of the salt solution, and the salt will have no effect. The effectiveness of a de-icing salt depends on the number of particles it releases on dissociation and on its solubility in water:
| name | Formula | lowest practical T, °C |
|---|---|---|
| ammonium sulfate | (NH 4 ) 2 SO 4 | –7 |
| calcium chloride | CaCl 2 | –29 |
| potassium chloride | KCl | –15 |
| sodium chloride | NaCl | –9 |
| urea | (NH 2 ) 2 CO | –7 |
Automotive radiator antifreezes are mostly based on ethylene glycol, (CH 2 OH) 2 . Owing to the strong hydrogen-bonding properties of this double alcohol, this substance is miscible with water in all proportions, and contributes only a very small vapor pressure of its own. Besides lowering the freezing point, antifreeze also raises the boiling point, increasing the operating range of the cooling system. The pure glycol freezes at –12.9°C and boils at 197°C, allowing water-glycol mixtures to be tailored to a wide range of conditions.
Estimate the freezing point of an antifreeze mixture is made up by combining one volume of ethylene glycol (MW = 62, density 1.11 g cm –3 ) with two volumes of water.
Solution
Assume that we use 1 L of glycol and 2 L of water (the actual volumes do not matter as long as their ratios are as given.) The mass of the glycol will be 1.10 kg and that of the water will be 2.0 kg, so the total mass of the solution is 3.11 kg. We then have:
- number of moles of glycol: (1110 g) ÷ (62 g mol –1 ) = 17.9 mol
- molality of glycol: (17.9 mol) ÷ (2.00 kg) = 8.95 mol kg –1
- freezing point depression: Δ T F = (–1.86 K kg –1 mol) × (8.95 mol kg –1 ) = –16.6 K so the solution will freeze at about –17°C.
Any ionic species formed by dissociation will also contribute to the freezing point depression. This can serve as a useful means of determining the fraction of a solute that is dissociated.
An aqueous solution of nitrous acid (HNO 2 , MW = 47) freezes at –0.198 .C. If the solution was prepared by adding 0.100 mole of the acid to 1000 g of water, what percentage of the HNO 2 is dissociated in the solution?
Solution
The nominal molality of the solution is (.001 mol) ÷ (1.00 kg) = 0.001 mol kg –1 .
But the effective molality according to the observed Δ T F value is given by
Δ T F ÷ K F = (–.198 K) ÷(–1.86 K kg mol –1 ) = 0.106 mol kg –1 ; this is the total number of moles of species present after the dissociation reaction HNO 2 → H + + NO – has occurred. If we let x = [H + ] = [NO 2 – ], then by stoichiometry, [HNO 2 ] = 0.100 - x and .106 - x = 2x and x = .0355. The fraction of HNO 2 that is dissociated is .0355 ÷ 0.100 = .355, corresponding to 35.5% dissociation of the acid.
Another Perspective of Freezing Point Depression and Boiling Point Elevation
A simple phase diagram can provide more insight into these phenomena. You may already be familiar with the phase map for water below.
The one shown below expands on this by plotting lines for both pure water and for its "diluted" state produced by the introduction of a non-volatile solute.
The normal boiling point of the pure solvent is indicated by point where the vapor pressure curve intersects the 1-atm line — that is, where the escaping tendency of solvent molecules from the liquid is equivalent to 1 atmosphere pressure. Addition of a non-volatile solute reduces the vapor pressures to the values given by the blue line. This shifts the boiling point to the right , corresponding to the increase in temperature Δ T b required to raise the escaping tendency of the H 2 O molecules back up to 1 atm.
To understand freezing point depression, notice that the vapor pressure line intersects the curved black vapor pressure line of the solid (ice), which corresponds to a new triple point at which all three phases (ice, water vapor, and liquid water) are in equilibrium and thus exhibit equal escaping tendencies. This point is by definition the origin of the freezing (solid-liquid) line, which intersects the 1-atm line at a reduced freezing point Δ T f , indicated by .
Note that the above analysis assumes that the solute is soluble only in the liquid solvent, but not in its solid form. This is generally more or less true. For example, when arctic ice forms from seawater, the salts get mostly "squeezed" out. This has the interesting effect of making the water that remains more saline, and hence more dense, causing it to sink to the bottom part of the ocean where it gets taken up by the south-flowing deep current.
A Thermodynamics Perspective on Freezing and Boiling
Those readers who have some knowledge of thermodynamics will recognize that what we have been referring to as "escaping" tendency is really a manifestation of the Gibbs Energy . This schematic plot shows how the G 's for the solid, liquid, and gas phases of a typical substance vary with the temperature.
The rule is that the phase with the most negative free energy rules.
The phase that is most stable (and which therefore is the only one that exists) is always the one having the most negative free energy (indicated here by the thicker portions of the plotted lines.) The melting and boiling points correspond to the respective temperatures where the solid and liquid and liquid and vapor have identical free energies.
As we saw above, adding a solute to the liquid dilutes it, making its free energy more negative, with the result that the freezing and boiling points are shifted to the left and right, respectively.
The relationships shown in these plots depend on the differing slopes of the lines representing the free energies of the phases as the temperature changes. These slopes are proportional to the entropy of each phase. Because gases have the highest entropies, the slope of the "gaseous solvent" line is much greater than that of the others. Note that this plot is not to scale.


11.1: Fatty Acids
- To recognize the structures of common fatty acids and classify them as saturated, monounsaturated, or polyunsaturated.
Fatty acids are carboxylic acids that are structural components of fats and oils. They usually contain an even number of carbon atoms (typically 12–20), and can be classified by the presence and number of carbon-to-carbon double bonds. Note the presence of a nonpolar hydrocarbon chain (tail) and a polar carboxylic acid functional group (head) in the molecule. A fatty acid with one carbon-carbon double bond is a monounsaturated fatty acid. A fatty acid with two or more double bonds is a polyunsaturated fatty acid.
Figure \(\PageIndex{1}\): Saturated fatty acids have only single bonds while monounsaturated have one double bond and polyunsaturated have more than one double bond.
Saturated Fatty Acids
S aturated fatty acids contain all carbon-carbon single bonds. This causes the molecules to form straight chains, as shown in the figure 11.1.2. Palmitic acid is an example of a saturated fatty acid. The hydrocarbon tails of saturated fatty acids are able to be packed together very tightly with one another, interacting through London forces. London forces depends on the surface area of the molecule. The longer the hydrocarbon tail in a fatty acid, the stronger the interactions between molecules. The melting points of saturated fatty acids increases as the molecules get larger. This explains why saturated fatty acids are solids at room temperature. Animals use saturated fatty acids to store energy.
Unsaturated Fatty Acids
In an u nsa turated fatty acids , the hydrocarbon chain contains at least one carbon-carbon double bond (alkene). The double bonds are usually cis. The presence of a cis double bond causes chains to bend or kink in the hydrocarbon tail (see figure 11.1.2). The bent tails in neighboring unsaturated fatty acids are farther apart from one another compared to saturated fatty acids. This reduction of contact between the tails weakens the london forces between them. So unsaturated fatty acids are liquids at room temperature. For fatty acids with the same number of carbon atoms, the presence of more double bonds lowers the melting point. Plants use unsaturated fatty acids to store energy.
Figure \(\PageIndex{2}\): Structures of saturated and unsaturated fatty acids.
Essential Fatty Acids
Two polyunsaturated fatty acids—linoleic and linolenic acids—are termed essential fatty acids because humans must obtain them from their diets. Both substances are required for normal growth and development, but the human body does not synthesize them. The average daily diet should contain about 4–6 g of the essential fatty acids.Common Fatty Acids
Table \(\PageIndex{1}\) lists some common fatty acids. Chemists often use a shorthand notation for fatty acids along with the common name because they are all carboxylic acids with different numbers of carbons and number of double bonds. This notation gives the number of carbons followed by the number of double bonds present. For example: palmitic acid is a 16-carbon saturated fatty acid, would be represented by C16:0 and palmitoleic acid is a 16-carbon monounsaturated fatty acid, would be C16:1.| Name | Number of Carbons | Number of Double Bonds | Condensed Structural Formula | Melting Point (°C) | Shorthand Notation | Source |
|---|---|---|---|---|---|---|
| lauric acid | 12 | 0 | CH 3 (CH 2 ) 10 COOH | 44 | C12:0 | palm kernel oil |
| myristic acid | 14 | 0 | CH 3 (CH 2 ) 12 COOH | 58 | C14:0 | oil of nutmeg |
| palmitic acid | 16 | 0 | CH 3 (CH 2 ) 14 COOH | 63 | C16:0 | palm oil |
| palmitoleic acid | 16 | 1 | CH 3 (CH 2 ) 5 CH=CH(CH 2 ) 7 COOH | 0.5 | C16:1 | macadamia oil |
| stearic acid | 18 | 0 | CH 3 (CH 2 ) 16 COOH | 70 | C18:0 | cocoa butter |
| oleic acid | 18 | 1 | CH 3 (CH 2 ) 7 CH=CH(CH 2 ) 7 COOH | 16 | C18:1 | olive oil |
| linoleic acid | 18 | 2 | CH 3 (CH 2 ) 3 (CH 2 CH=CH) 2 (CH 2 ) 7 COOH | −5 | C18:2 | canola oil |
| linolenic acid | 18 | 3 | CH 3 (CH 2 CH=CH) 3 (CH 2 ) 7 COOH | −11 | C18:3 | flaxseed |
| arachidonic acid | 20 | 4 | CH 3 (CH 2 ) 4 (CH=CHCH 2 ) 4 (CH 2 ) 2 COOH | −50 | C20:4 | liver |
ω-3 Fatty acids
Fats containing saturated fatty acids are considered not so healthy as they have a tendency to deposit in arteries forming plaque, i.e., atherosclerosis. It can lead to high blood pressure, rupture of arteries, or heart attack. Saturated fatty acids are more common in animal fats while unsaturated fatty acids which are considered more healthy are more common in vegetable oils.
In omega (ω) labeling the last C, i.e., the −CH 3 at the end opposite to carboxyl (−COOH) group is labeled ω-1, the C next to it as ω-2, then ω-3 and so on, as as shown in the figure 11.1.3. Omega (ω) is the last alphabet in Greek, so, ω-C is the last carbon. The (ω) label is used to refer to the position of the first C=C bond from the ω-end. For example, linolenic acid in figure below is an ω-3 fatty acid.
Figure \(\PageIndex{3}\): Labelling of C′s in fatty acids: systematic IUPAC numbers are in red, nonsystematic Greek letter labels are in black, and omega- (ω-#) are in blue. (Copyright; Public domain)
It has been observed that peoples from Alaska eat a higher amount of unsaturated fats and have a low occurrence of atherosclerosis and heart attack. Fats in Alaskan food is primarily from fish, such as salmon, tuna, and herring. Vegetable oils have higher contents of ω-6 fatty acids, such as linoleic acid and arachidonic acid. Fish oil mostly contains ω-3 fatty acids, such as linolenic acid, eicosapentaenoic acid, and docosahexaenoic acid.
Fatty Acid Compositions of Common Fats and Oils
Table 11.1.2 shows the fatty acid compositions of some common fats and oils. The composition of any given fat or oil can vary depending on the plant or animal species it comes from as well as on dietetic and climatic factors. To cite just one example, lard from corn-fed hogs is more highly saturated than lard from peanut-fed hogs.  Terms such as saturated fat or unsaturated oil are often used to describe the fats or oils obtained from foods. Saturated fats contain a high proportion of saturated fatty acids, while unsaturated oils contain a high proportion of unsaturated fatty acids. The high consumption of saturated fats is a factor, along with the high consumption of cholesterol, in increased risks of heart disease.| Source | Lauric (C12:0) | Myristic (C14:0) | Palmitic (C16:0) | Stearic (C18:0) | Oleic (C18:1) | Linoleic (C18:2) | Linolenic (C18:3) |
|---|---|---|---|---|---|---|---|
| Fats | |||||||
| butter (cow) | 3 | 11 | 27 | 12 | 29 | 2 | 1 |
| tallow | 3 | 24 | 19 | 43 | 3 | 1 | |
| lard | 2 | 26 | 14 | 44 | 10 | ||
| Oils | |||||||
| canola oil | 4 | 2 | 62 | 22 | 10 | ||
| coconut oil † | 47 | 18 | 9 | 3 | 6 | 2 | |
| corn oil | 11 | 2 | 28 | 58 | 1 | ||
| olive oil | 13 | 3 | 71 | 10 | 1 | ||
| peanut oil | 11 | 2 | 48 | 32 | |||
| soybean oil | 11 | 4 | 24 | 54 | 7 | ||
| *Totals less than 100% indicate the presence of fatty acids with fewer than 12 carbon atoms or more than 18 carbon atoms. | |||||||
| † Coconut oil is highly saturated. It contains an unusually high percentage of the low-melting C 8 , C 10 , and C 12 saturated fatty acids. |


17.5: Inflammation and Fever
- Identify the signs of inflammation and fever and explain why they occur
- Explain the advantages and risks posed by inflammatory responses
The inflammatory response, or inflammation, is triggered by a cascade of chemical mediators and cellular responses that may occur when cells are damaged and stressed or when pathogens successfully breach the physical barriers of the innate immune system. Although inflammation is typically associated with negative consequences of injury or disease, it is a necessary process insofar as it allows for recruitment of the cellular defenses needed to eliminate pathogens, remove damaged and dead cells, and initiate repair mechanisms. Excessive inflammation, however, can result in local tissue damage and, in severe cases, may even become deadly.
Acute Inflammation
An early, if not immediate, response to tissue injury is acute inflammation. Immediately following an injury, vasoconstriction of blood vessels will occur to minimize blood loss. The amount of vasoconstriction is related to the amount of vascular injury, but it is usually brief. Vasoconstriction is followed by vasodilation and increased vascular permeability, as a direct result of the release of histamine from resident mast cells. Increased blood flow and vascular permeability can dilute toxins and bacterial products at the site of injury or infection. They also contribute to the five observable signs associated with the inflammatory response: erythema (redness), edema (swelling), heat, pain, and altered function. Vasodilation and increased vascular permeability are also associated with an influx of phagocytes at the site of injury and/or infection. This can enhance the inflammatory response because phagocytes may release proinflammatory chemicals when they are activated by cellular distress signals released from damaged cells, by PAMPs, or by opsonins on the surface of pathogens. Activation of the complement system can further enhance the inflammatory response through the production of the anaphylatoxin C5a. Figure \(\PageIndex{1}\) illustrates a typical case of acute inflammation at the site of a skin wound.
During the period of inflammation, the release of bradykinin causes capillaries to remain dilated, flooding tissues with fluids and leading to edema. Increasing numbers of neutrophils are recruited to the area to fight pathogens. As the fight rages on, pus forms from the accumulation of neutrophils, dead cells, tissue fluids, and lymph. Typically, after a few days, macrophages will help to clear out this pus. Eventually, tissue repair can begin in the wounded area.
Chronic Inflammation
When acute inflammation is unable to clear an infectious pathogen, chronic inflammation may occur. This often results in an ongoing (and sometimes futile) lower-level battle between the host organism and the pathogen. The wounded area may heal at a superficial level, but pathogens may still be present in deeper tissues, stimulating ongoing inflammation. Additionally, chronic inflammation may be involved in the progression of degenerative neurological diseases such as Alzheimer’s and Parkinson’s, heart disease, and metastatic cancer.
Chronic inflammation may lead to the formation of granulomas, pockets of infected tissue walled off and surrounded by WBCs. Macrophages and other phagocytes wage an unsuccessful battle to eliminate the pathogens and dead cellular materials within a granuloma. One example of a disease that produces chronic inflammation is tuberculosis, which results in the formation of granulomas in lung tissues. A tubercular granuloma is called a tubercle (Figure \(\PageIndex{2}\)). Tuberculosis will be covered in more detail in Bacterial Infections of the Respiratory Tract .
Chronic inflammation is not just associated with bacterial infections. Chronic inflammation can be an important cause of tissue damage from viral infections. The extensive scarring observed with hepatitis C infections and liver cirrhosis is the result of chronic inflammation.
- Name the five signs of inflammation.
- Is a granuloma an acute or chronic form of inflammation? Explain.
In addition to granulomas, chronic inflammation can also result in long-term edema. A condition known as lymphatic filariasis (also known as elephantiasis) provides an extreme example. Lymphatic filariasis is caused by microscopic nematodes (parasitic worms) whose larvae are transmitted between human hosts by mosquitoes. Adult worms live in the lymphatic vessels, where their presence stimulates infiltration by lymphocytes, plasma cells, eosinophils, and thrombocytes (a condition known as lymphangitis). Because of the chronic nature of the illness, granulomas, fibrosis, and blocking of the lymphatic system may eventually occur. Over time, these blockages may worsen with repeated infections over decades, leading to skin thickened with edema and fibrosis. Lymph (extracellular tissue fluid) may spill out of the lymphatic areas and back into tissues, causing extreme swelling (Figure \(\PageIndex{3}\)). Secondary bacterial infections commonly follow. Because it is a disease caused by a parasite, eosinophilia (a dramatic rise in the number of eosinophils in the blood) is characteristic of acute infection. However, this increase in antiparasite granulocytes is not sufficient to clear the infection in many cases.
Lymphatic filariasis affects an estimated 120 million people worldwide, mostly concentrated in Africa and Asia. 1 Improved sanitation and mosquito control can reduce transmission rates.
Fever
A fever is an inflammatory response that extends beyond the site of infection and affects the entire body, resulting in an overall increase in body temperature. Body temperature is normally regulated and maintained by the hypothalamus, an anatomical section of the brain that functions to maintain homeostasis in the body. However, certain bacterial or viral infections can result in the production of pyrogens, chemicals that effectively alter the “thermostat setting” of the hypothalamus to elevate body temperature and cause fever. Pyrogens may be exogenous or endogenous. For example, the endotoxin lipopolysaccharide (LPS), produced by gram-negative bacteria, is an exogenous pyrogen that may induce the leukocytes to release endogenous pyrogens such as interleukin-1 (IL-1), IL-6, interferon-γ (IFN-γ), and tumor necrosis factor (TNF). In a cascading effect, these molecules can then lead to the release of prostaglandin E2 (PGE 2 ) from other cells, resetting the hypothalamus to initiate fever (Figure \(\PageIndex{4}\)).
Like other forms of inflammation, a fever enhances the innate immune defenses by stimulating leukocytes to kill pathogens. The rise in body temperature also may inhibit the growth of many pathogens since human pathogens are mesophiles with optimum growth occurring around 35 °C (95 °F). In addition, some studies suggest that fever may also stimulate release of iron-sequestering compounds from the liver, thereby starving out microbes that rely on iron for growth. 2
During fever, the skin may appear pale due to vasoconstriction of the blood vessels in the skin, which is mediated by the hypothalamus to divert blood flow away from extremities, minimizing the loss of heat and raising the core temperature. The hypothalamus will also stimulate shivering of muscles, another effective mechanism of generating heat and raising the core temperature.
The crisis phase occurs when the fever breaks. The hypothalamus stimulates vasodilation, resulting in a return of blood flow to the skin and a subsequent release of heat from the body. The hypothalamus also stimulates sweating, which cools the skin as the sweat evaporates.
Although a low-level fever may help an individual overcome an illness, in some instances, this immune response can be too strong, causing tissue and organ damage and, in severe cases, even death. The inflammatory response to bacterial superantigens is one scenario in which a life-threatening fever may develop. Superantigens are bacterial or viral proteins that can cause an excessive activation of T cells from the specific adaptive immune defense, as well as an excessive release of cytokines that overstimulates the inflammatory response. For example, Staphylococcus aureus and Streptococcus pyogenes are capable of producing superantigens that cause toxic shock syndrome and scarlet fever, respectively. Both of these conditions can be associated with very high, life-threatening fevers in excess of 42 °C (108 °F).
- Explain the difference between exogenous and endogenous pyrogens.
- How does a fever inhibit pathogens?
Given her father’s premature death, Angela’s doctor suspects that she has hereditary angioedema, a genetic disorder that compromises the function of C1 inhibitor protein. Patients with this genetic abnormality may have occasional episodes of swelling in various parts of the body. In Angela’s case, the swelling has occurred in the respiratory tract, leading to difficulty breathing. Swelling may also occur in the gastrointestinal tract, causing abdominal cramping, diarrhea, and vomiting, or in the muscles of the face or limbs. This swelling may be nonresponsive to steroid treatment and is often misdiagnosed as an allergy.
Because there are three types of hereditary angioedema, the doctor orders a more specific blood test to look for levels of C1-INH, as well as a functional assay of Angela’s C1 inhibitors. The results suggest that Angela has type I hereditary angioedema, which accounts for 80%–85% of all cases. This form of the disorder is caused by a deficiency in C1 esterase inhibitors, the proteins that normally help suppress activation of the complement system. When these proteins are deficient or nonfunctional, overstimulation of the system can lead to production of inflammatory anaphylatoxins, which results in swelling and fluid buildup in tissues.
There is no cure for hereditary angioedema, but timely treatment with purified and concentrated C1-INH from blood donors can be effective, preventing tragic outcomes like the one suffered by Angela’s father. A number of therapeutic drugs, either currently approved or in late-stage human trials, may also be considered as options for treatment in the near future. These drugs work by inhibiting inflammatory molecules or the receptors for inflammatory molecules.
Thankfully, Angela’s condition was quickly diagnosed and treated. Although she may experience additional episodes in the future, her prognosis is good and she can expect to live a relatively normal life provided she seeks treatment at the onset of symptoms.
Key Concepts and Summary
- Inflammation results from the collective response of chemical mediators and cellular defenses to an injury or infection.
- Acute inflammation is short lived and localized to the site of injury or infection. Chronic inflammation occurs when the inflammatory response is unsuccessful, and may result in the formation of granulomas (e.g., with tuberculosis) and scarring (e.g., with hepatitis C viral infections and liver cirrhosis).
- The five cardinal signs of inflammation are erythema , edema , heat, pain, and altered function. These largely result from innate responses that draw increased blood flow to the injured or infected tissue.
- Fever is a system-wide sign of inflammation that raises the body temperature and stimulates the immune response.
- Both inflammation and fever can be harmful if the inflammatory response is too severe.
Footnotes
- 1 Centers for Disease Control and Prevention. “Parasites–Lymphatic Filiariasis.” 2016. http://www.cdc.gov/parasites/lymphat...info/faqs.html .
- 2 N. Parrow et al. “Sequestration and Scavenging of Iron in Infection.” Infection and Immunity 81 no. 10 (2013):3503–3514


14.3: Arpeggiated Accompaniments
14.3 Arpeggiated Accompaniments
14.3.1 Arpeggios
One way to express chords rhythmically is through arpeggios in one part and a bass line in octaves in a lower part, as in the following example from Beethoven’s Moonlight Sonata.
The next example has descending arpeggios.
Notice that in both the Beethoven and Alicia Keys examples there is the harmonious interval of a tenth (an octave plus a third) between the bass (lowest) voice and the soprano (highest) voice. The following examples have arpeggios that ascend and descend through a chord.
Notice in the above example that there is also an organ playing block chords to create a sense of legato in the texture. The next two examples are from more recent popular music.
Below is an example in 44 with arpeggios in sixteenth notes.
14.3.2 Alberti Bass
Alberti bass accompaniment patterns involve arpeggios that do not arpeggiate chords in a simple upward or downward motion, but in a “low–high–middle–high” pattern as you can see in the examples below.
The next example uses the same Alberti pattern as in the Mozart example above, but transposed to E minor and in a lower register.


4.5: Electrical Potential Due to a Point Charge
Learning Objectives
By the end of this section, you will be able to:
- Explain point charges and express the equation for electric potential of a point charge.
- Distinguish between electric potential and electric field.
- Determine the electric potential of a point charge given charge and distance.
Point charges, such as electrons, are among the fundamental building blocks of matter. Furthermore, spherical charge distributions (like on a metal sphere) create external electric fields exactly like a point charge. The electric potential due to a point charge is, thus, a case we need to consider. Using calculus to find the work needed to move a test charge \(q\) from a large distance away to a distance of \(r\) from a point charge \(Q\), and noting the connection between work and potential \((W=-q\Delta V)\), we can define the electric potential \(V\) of a point charge:
definition: ELECTRIC POTENTIAL \(V\) OF A POINT CHARGE
The electric potential \(V\) of a point charge is given by
\[V=\dfrac{kQ}{r}\: (\mathrm{Point\: Charge}). \label{eq1}\]
where \(k\) is a constant equal to \(9.0 \times 10^{9}\, \mathrm{N}\cdot \mathrm{m^{2}/C^{2}}.\)
The potential at infinity is chosen to be zero. Thus \(V\) for a point charge decreases with distance, whereas \(\mathbf{E}\) for a point charge decreases with distance squared:
\[E=\dfrac{F}{q}=\dfrac{kQ}{r^{2}}.\]
Recall that the electric potential \(V\) is a scalar and has no direction, whereas the electric field \(\mathbf{E}\) is a vector. To find the voltage due to a combination of point charges, you add the individual voltages as numbers. To find the total electric field, you must add the individual fields as vectors , taking magnitude and direction into account. This is consistent with the fact that \(V\) is closely associated with energy, a scalar, whereas \(\mathbf{E}\) is closely associated with force, a vector.
Example \(\PageIndex{1}\): What Voltage Is Produced by a Small Charge on a Metal Sphere?
Charges in static electricity are typically in the nanocoulomb \((\mathrm{nC})\) to microcoulomb \((\mu \mathrm{C})\) range. What is the voltage 5.00 cm away from the center of a 1-cm diameter metal sphere that has a \(-3.00 \mathrm{nC}\) static charge?
Strategy
As we have discussed in Electric Charge and Electric Field , charge on a metal sphere spreads out uniformly and produces a field like that of a point charge located at its center. Thus we can find the voltage using Equation \ref{eq1}.
Solution
Entering known values into the expression for the potential of a point charge, we obtain
\[ \begin{align*} V&=k\dfrac{Q}{r} \\[5pt] &=(8.99 \times 10^{9} \, \mathrm{N}\cdot \mathrm{m^{2}/C^{2}}) \left(\dfrac{-3.00\times 10^{-9}\,\mathrm{C}}{5.00\times 10^{-2}\,\mathrm{m}}\right) \\[5pt] &= -539\, \mathrm{V}. \end{align*}\]
Discussion
The negative value for voltage means a positive charge would be attracted from a larger distance, since the potential is lower (more negative) than at larger distances. Conversely, a negative charge would be repelled, as expected.
Example \(\PageIndex{2}\): What Is the Excess Charge on a Van de Graaff Generator
A demonstration Van de Graaff generator has a 25.0 cm diameter metal sphere that produces a voltage of 100 kV near its surface. (Figure \(\PageIndex{1}\)) What excess charge resides on the sphere? (Assume that each numerical value here is shown with three significant figures.)
Strategy
The potential on the surface will be the same as that of a point charge at the center of the sphere, 12.5 cm away. (The radius of the sphere is 12.5 cm.) We can thus determine the excess charge using Equation \ref{eq1}.
Solution
Solving for \(Q\) and entering known values gives
\[ \begin{align*} Q &=\dfrac{rV}{k} \\[5pt] &= \dfrac{(0.125 \,\mathrm{m})(100\times 10^{3}\, \mathrm{V})}{8.99\times 10^{9}\, \mathrm{N\cdot m^{2}/C^{2}}} \\[5pt] &= 1.39\times 10^{-6} \,\mathrm{C} \\[5pt] &= 1.39\, \mathrm{\mu C}.\end{align*}\]
Discussion
This is a relatively small charge, but it produces a rather large voltage. We have another indication here that it is difficult to store isolated charges.
The voltages in both of these examples could be measured with a meter that compares the measured potential with ground potential. Ground potential is often taken to be zero (instead of taking the potential at infinity to be zero). It is the potential difference between two points that is of importance, and very often there is a tacit assumption that some reference point, such as Earth or a very distant point, is at zero potential. This is analogous to taking sea level as \(h=0\) when considering gravitational potential energy, \(\mathrm{PE_{g}}=mgh\).
Consider the dipole in Figure \(\PageIndex{3}\) with the charge magnitude of \(q = 3.0 \, \mu C\) and separation distance \(d = 4.0 \, cm.\) What is the potential at the following locations in space? (a) (0, 0, 1.0 cm); (b) (0, 0, –5.0 cm); (c) (3.0 cm, 0, 2.0 cm).
Strategy
Apply \(V_p = k \sum_1^N \dfrac{q_i}{r_i}\) to each of these three points.
Solution
a. \(V_p = k \sum_1^N \dfrac{q_i}{r_i} = (9.0 \times 10^9 \, N \cdot m^2/C^2) \left(\dfrac{3.0\space nC}{0.010 \, m} - \dfrac{3.0\space nC}{0.030 \, m}\right) = 1.8 \times 10^3 \, V\)
b. \(V_p = k \sum_1^N \dfrac{q_i}{r_i} = (9.0 \times 10^9 \, N \cdot m^2/C^2) \left(\dfrac{3.0\space nC}{0.070 \, m} - \dfrac{3.0\space nC}{0.030 \, m}\right) = -5.1 \times 10^2 \, V\)
c. \(V_p = k \sum_1^N \dfrac{q_i}{r_i} = (9.0 \times 10^9 \, N \cdot m^2/C^2) \left(\dfrac{3.0\space nC}{0.030 \, m} - \dfrac{3.0\space nC}{0.050 \, m}\right) = 3.6 \times 10^2 \, V\)
Significance
Note that evaluating potential is significantly simpler than electric field, due to potential being a scalar instead of a vector.
Summary
- Electric potential of a point charge is \(V=kQ/r\).
- Electric potential is a scalar, and electric field is a vector. Addition of voltages as numbers gives the voltage due to a combination of point charges, whereas addition of individual fields as vectors gives the total electric field.


11.1: Introduction
Religious practitioners around the world have long used music to express their spiritual convictions. Music plays a particularly significant role in the context of worship. This is hardly surprising: Music has the power to encourage a wide variety of emotional states, ranging from meditative calm to frenzied excitement. It can help participants in worship to feel close to each other and to their deity. It also aids in the memorization and communal recitation of texts, which often define religious practice. Individual creative artists also draw inspiration from their religious convictions, even if the music they produce is not intended for the purpose of facilitating worship.
In this chapter, we will explore a variety of examples related to Christian worship and beliefs. The focus on Christianity is the result of the Western and classical bias of this text. Christianity has been the dominant religion of Europe since the 4th century, when it was legalized by the Roman emperor Constantine the Great. Most of the influential composers in the classical tradition belonged to Christian denominations, and many wrote beautiful music for use in church services. As a result, church music through the ages has both reflected and shaped broader musical practices. The stylistic variety of this music is nearly infinite. This variety results from the combined forces of general musical taste, the requirements of religious authorities, and the needs and histories of congregations.
This chapter will be dedicated to the examination of worship music from different eras. Although today we are more likely to hear most of this music in concert halls or on recordings, much of it was first intended for practical use in church services. To understand how this music came to be and what it meant in its time and place, we will consider each example in its religious context.


7.3: Synapses
- Explain how sample signals cross the synapse
Synaptic Transmission
The synapse or “gap” is the place where information is transmitted from one neuron to another. Synapses usually form between axon terminals and dendritic spines, but this is not universally true. There are also axon-to-axon, dendrite-to-dendrite, and axon-to-cell body synapses. The neuron transmitting the signal is called the presynaptic neuron, and the neuron receiving the signal is called the postsynaptic neuron. Note that these designations are relative to a particular synapse—most neurons are both presynaptic and postsynaptic. There are two types of synapses: chemical and electrical.
Chemical Synapses
When an action potential reaches the axon terminal it depolarizes the membrane and opens voltage-gated \(\ce{Na^{+}}\) channels. \(\ce{Na^{+}}\) ions enter the cell, further depolarizing the presynaptic membrane. This depolarization causes voltage-gated \(\ce{Ca^{3+}}\) channels to open. Calcium ions entering the cell initiate a signaling cascade that causes small membrane-bound vesicles, called synaptic vesicles, containing neurotransmitter molecules to fuse with the presynaptic membrane. Synaptic vesicles are shown in Figure 7.18, which is an image from a scanning electron microscope.
Fusion of a vesicle with the presynaptic membrane causes the neurotransmitter to be released into the synaptic cleft, the extracellular space between the presynaptic and postsynaptic membranes, as illustrated in Figure 7.19 . The neurotransmitter diffuses across the synaptic cleft and binds to receptor proteins on the postsynaptic membrane.
The binding of a specific neurotransmitter causes particular ion channels, in this case, ligand-gated channels, on the postsynaptic membrane to open. Neurotransmitters can either have excitatory or inhibitory effects on the postsynaptic membrane, as detailed in Table 7.2. For example, when acetylcholine is released at the synapse between a nerve and muscle (called the neuromuscular junction) by a presynaptic neuron, it causes postsynaptic Na+ channels to open. Na+ enters the postsynaptic cell and causes the postsynaptic membrane to depolarize. This depolarization is called an excitatory postsynaptic potential (EPSP) and makes the postsynaptic neuron more likely to fire an action potential. The release of neurotransmitter at inhibitory synapses causes inhibitory postsynaptic potentials (IPSPs) , a hyperpolarization of the presynaptic membrane. For example, when the neurotransmitter GABA (gamma-aminobutyric acid) is released from a presynaptic neuron, it binds to and opens Cl– channels. Cl– ions enter the cell and hyperpolarizes the membrane, making the neuron less likely to fire an action potential.
Once neurotransmission has occurred, the neurotransmitter must be removed from the synaptic cleft so that the postsynaptic membrane can “reset” and be ready to receive another signal. This can be accomplished in three ways: the neurotransmitter can diffuse away from the synaptic cleft, it can be degraded by enzymes in the synaptic cleft, or it can be recycled (sometimes called reuptake) by the presynaptic neuron. Several drugs act at this step of neurotransmission. For example, some drugs that are given to Alzheimer’s patients work by inhibiting acetylcholinesterase, the enzyme that degrades acetylcholine. This inhibition of the enzyme essentially increases neurotransmission at synapses that release acetylcholine. Once released, the acetylcholine stays in the cleft and can continually bind and unbind to postsynaptic receptors.
| Neurotransmitter | Example | Location |
|---|---|---|
| Acetylcholine | — | CNS and/or PNS |
| Biogenic amine | Dopamine, serotonin, norepinephrine | CNS and/or PNS |
| Amino acid | Glycine, glutamate, aspartate, gamma-aminobutyric acid | CNS |
| Neuropeptide | Substance P, endorphins | CNS and/or PNS |
Design and/or act out events at the synapse! Here are some hints to help you design your review activity:
- You will need to think about how many peers you need to participate.
- Who will be which structure?
- How will your peers and you move with respect to each other?
- What is the overall goal of each movement you design (what is the function of each structure)?
Electrical Synapses
While electrical synapses are fewer in number than chemical synapses, they are found in all nervous systems and play important and unique roles. The mode of neurotransmission in electrical synapses is quite different from that in chemical synapses. In an electrical synapse, the presynaptic and postsynaptic membranes are very close together and are actually physically connected by channel proteins forming gap junctions. Gap junctions allow current to pass directly from one cell to the next. In addition to the ions that carry this current, other molecules, such as ATP, can diffuse through the large gap junction pores.
There are key differences between chemical and electrical synapses. Because chemical synapses depend on the release of neurotransmitter molecules from synaptic vesicles to pass on their signal, there is an approximately one-millisecond delay between when the axon potential reaches the presynaptic terminal and when the neurotransmitter leads to opening of postsynaptic ion channels. Additionally, this signaling is unidirectional. Signaling in electrical synapses, in contrast, is virtually instantaneous (which is important for synapses involved in key reflexes), and some electrical synapses are bidirectional. Electrical synapses are also more reliable as they are less likely to be blocked, and they are important for synchronizing the electrical activity of a group of neurons. For example, electrical synapses in the thalamus are thought to regulate slow-wave sleep, and disruption of these synapses can cause seizures.
Signal summation
Sometimes a single EPSP is strong enough to induce an action potential in the postsynaptic neuron, but often multiple presynaptic inputs must create EPSPs around the same time for the postsynaptic neuron to be sufficiently depolarized to fire an action potential. This process is called summation and occurs at the axon hillock, as illustrated in Figure 7.20. Additionally, one neuron often has inputs from many presynaptic neurons—some excitatory and some inhibitory—so IPSPs can cancel out EPSPs and vice versa. It is the net change in postsynaptic membrane voltage that determines whether the postsynaptic cell has reached its threshold of excitation needed to fire an action potential. Together, synaptic summation and the threshold for excitation act as a filter so that random “noise” in the system is not transmitted as important information.
Synaptic plasticity
Synapses are not static structures. They can be weakened or strengthened. They can be broken, and new synapses can be made. Synaptic plasticity allows for these changes, which are all needed for a functioning nervous system. In fact, synaptic plasticity is the basis of learning and memory. Two processes in particular, long-term potentiation (LTP) and long-term depression (LTD) are important forms of synaptic plasticity that occur in synapses in the hippocampus, a brain region that is involved in storing memories.
Long-term potentiation (LTP)
Long-term potentiation (LTP) is a persistent strengthening of a synaptic connection. LTP is based on the Hebbian principle: cells that fire together wire together. There are various mechanisms, none fully understood, behind the synaptic strengthening seen with LTP. One known mechanism involves a type of postsynaptic glutamate receptor, called NMDA (N-Methyl-D-aspartate) receptors, shown in Figure 7.21. These receptors are normally blocked by magnesium ions; however, when the postsynaptic neuron is depolarized by multiple presynaptic inputs in quick succession (either from one neuron or multiple neurons), the magnesium ions are forced out allowing Ca ions to pass into the postsynaptic cell. Next, Ca2+ions entering the cell initiate a signaling cascade that causes a different type of glutamate receptor, called AMPA (α-amino-3-hydroxy-5-methyl-4-isoxazolepropionic acid) receptors, to be inserted into the postsynaptic membrane, since activated AMPA receptors allow positive ions to enter the cell. So, the next time glutamate is released from the presynaptic membrane, it will have a larger excitatory effect (EPSP) on the postsynaptic cell because the binding of glutamate to these AMPA receptors will allow more positive ions into the cell. The insertion of additional AMPA receptors strengthens the synapse and means that the postsynaptic neuron is more likely to fire in response to presynaptic neurotransmitter release. Some drugs of abuse co-opt the LTP pathway, and this synaptic strengthening can lead to addiction.
Long-term depression (LTD)
Long-term depression (LTD) is essentially the reverse of LTP: it is a long-term weakening of a synaptic connection. One mechanism known to cause LTD also involves AMPA receptors. In this situation, calcium that enters through NMDA receptors initiates a different signaling cascade, which results in the removal of AMPA receptors from the postsynaptic membrane, as illustrated in Figure 7.21. The decrease in AMPA receptors in the membrane makes the postsynaptic neuron less responsive to glutamate released from the presynaptic neuron. While it may seem counterintuitive, LTD may be just as important for learning and memory as LTP. The weakening and pruning of unused synapses allow for unimportant connections to be lost and makes the synapses that have undergone LTP that much stronger by comparison.
For a neuron to fire an action potential, its membrane must reach ________.
a. hyperpolarization
b. the threshold of excitation
c. the refractory period
d. inhibitory postsynaptic potentia
After an action potential, the opening of additional voltage-gated ________ channels and the inactivation of sodium channels, cause the membrane to return to its resting membrane potential.
a. sodium
b. potassium
c. calcium
d. chloride
What is the term for protein channels that connect two neurons at an electrical synapse?
a. synaptic vesicles
b. voltage-gated ion channels
c. gap junction protein
d. sodium-potassium exchange pumps
How does myelin aid propagation of an action potential along an axon? How do the nodes of Ranvier help this process?
What are the main steps in chemical neurotransmission?